5

Use the Chain Rule to find the Indicated partial derivatives_2 = x xy2 ,* = UV2 w2,Y = U vewdu dv Jw when u = 2,V = 1,w = 0dudvDwNeed Help?Read IiWatch#tSubmit Answ...

Question

Use the Chain Rule to find the Indicated partial derivatives_2 = x xy2 ,* = UV2 w2,Y = U vewdu dv Jw when u = 2,V = 1,w = 0dudvDwNeed Help?Read IiWatch#tSubmit Answer19. [~/2 Points]DETAILSSCALCCC4 11.5.024.Use the Chain Rule to find the indicated partial derivatives_M = xey - 22 , X = Juv,Y = u V,2 = UOM DM du when u = l,v =-2OM duDM dv

Use the Chain Rule to find the Indicated partial derivatives_ 2 = x xy2 ,* = UV2 w2,Y = U vew du dv Jw when u = 2,V = 1,w = 0 du dv Dw Need Help? Read Ii Watch#t Submit Answer 19. [~/2 Points] DETAILS SCALCCC4 11.5.024. Use the Chain Rule to find the indicated partial derivatives_ M = xey - 22 , X = Juv,Y = u V,2 = U OM DM du when u = l,v =-2 OM du DM dv



Answers

$21-26$ Use the Chain Rule to find the indicated partial derivatives.
$$\begin{array}{l}{M=x e^{y-z^{2}}, \quad x=2 u v, \quad y=u-v, \quad z=u+v} \\ {\frac{\partial M}{\partial u}, \frac{\partial M}{\partial v} \text { when } u=3, v=-1}\end{array}$$

This problem, we are given that a functional R is a function of three values for three variables U V and W. And that U V and w are all functions of X and Y. As written in this chart right here. And we are asked to find the RdX and D r d Y. Now in this question to use this, we need to use the chain rule, but it's gonna be a little bit longer than we have in the past. And writing this chart on the left hand side is very, very efficient because it shows us that wherever we see an X here, here and here, we're going to have three X is at the bottom, which means that we have to have three different terms to add together. Because there are three elements of this. There is the you part, there's the V park and there's the W part. Okay. And so if we draw this out showing all of this here, we say that we're gonna have three components. The first one is D r d u multiplied by the U D X. We have the R d v times DVD X and we have D R D W time's D W T X. And this can just be exchanged very quickly for the Y2 because all of them are functions of why as well. And I just kind of want to highlight here that, that we are doing this, we should, you should always see that the, the use kind of try to cancel each other out in a way, right? They kind of cancel each other out. We should always get kind of that, you know that that D R D X form in every single term, it's a good way to kind of check yourself. So we just take the derivatives Now D R D the derivative are with respect to you here. Well, the derivative, the natural log is one over the argument. Multiply by the derivative of the inside. Right? Um so we have one divided by U squared plus B squared plus W squared. Have you squared multiplied by the derivative of the inside with respect to you, which is just two times you? Yeah. And then we have the U D X Which is just one. Now this is a very similar feel. Every single time with the derivative are with respect to be so it would be one overuse grip of speed script was W squared multiplied by this time. It's to be then DVD x in this case is just gonna be too, so not too bad. And lastly here, D r d w is going to just be two W divided by U squared plus B squared plus W squared and multiply by D W D X which is going to be equal to two times Y Yeah. Um there's a really, I guess there's a nice way you might be able to simplify this here if you wanted to. Um it's not necessary that you do because you can pull out the U squared plus b squared W. Skirt. I wouldn't do that. I think at this point you can just start plugging things in. So what you can do is you know that X. Is equal to y. It was a little one. We're trying to find the value at excellent like little one but we have a lot of these that are have values of U. V. And W. We can see that plugging in one to you would actually make this 1-plus 2 which is able to three for a deal for V. We would get here that this is equal to just 2 -1 which is equal to one. N. W. Would be equal to just two because we have two times one times one. So wherever we see you we plug in three wherever she will be one. W. is too. So we get That this is going to be six at the top. On the bottom will be three squared plus one squared plus two squared plus two times one divided by three squared plus one squared plus two squared Times two Times 2 and full plus. And we get two times two. This time have three squared plus one squared plus two squared. And then and then why here is equal to one. So we get two times one and we see that this is six divided by so nine square plus one squared plus four squared sorry nine plus one plus four is equal to 14. This is able to to over 14. Oh sorry, Two times two which is equal to 4/14. And lastly here we have two times you damn shoes, which is 8/14 and we get 18/14. This has this baseball can divide by two, So we can get something like 9/7. Mhm. And this is our answer and this is the exact same answer. Not exact same answer, but the same exact process you would do for D. R. D. Y. But do you argue why this part this part and this part do not change, right? These parts do not change because I'm going to have to take the D. R. D. U. And D R D V D R D W. That's that's possible. That's not changing at all. So we would again get, you know, this year? This year and this here the only things that change our that we have to compute now, D. U. D Dy which is equipped to DVDY which is equal to negative one and D. W. D. Y. Which is equal to two. X. Sound good. And so from there we would again get 6/14, multiplied by two Plus. This one would be to over 14. Sorry to over 14 divided by times negative one plus. In this case this would be 4/14 Times two times a which is just to Mhm. And so we get 12/14 -2 or 14 Plus 8/14. And we get that this is actually 18/14. And how about that? We get that. This is 97 again. And so this problem gives us an idea of how to use a function, take a derivative of a function now with three terms, and this extends very very nicely. So this is just important for you to remember. This extends over and over and over again. As we get more and more and more variables draw out this picture, it's extremely help.

Okay, F is a function of U. And V. And U and V. R. Functions of X and Y. So if we want to find diva D partial that with respect to why we gotta do D F d u d u D Y. Do you have t v DVD y ourself F with respect to you times the partial of U. With respect to why? Plus the partial of F. With respect to me times the partial viv? With respect to why? Alright, so F is E to the U plus V. So the derivative of E to the U plus V is E. To the U plus V Times the derivative of you plus B with respect to you. Which is one times the derivative of U. With respect to Y zero plus D. F T V. That's E to the U plus B times the derivative of U plus V. Which is one times T V Dy, which is X. So X. E. To the U plus B.

Okay. F is a function of X and Y. X and Y are both functions of U. And V. They want us to find the partial of F with respect to you. So we're gonna do DF dx dx d you plus D. F. Dy. Dy, Do you? So the derivative of F with respect to access to X times the derivative of X. With respect to you? That's E. To some power. So it's derivative is eat the power times the derivative of the power, which is just one plus Dfd Y. Which is to Y times the derivative of why with respect to you? Which is just one. So you get two X. E. To the U plus V plus two wild.

So for this problem we have to find the derivative of Z with respect to U. V. And W. Given that Z is a function of X. Y. And that X and Y are both functions of U. V and W. So we want to just remember that drawing out this diagram, this tree diagrams. Really helpful for us to figure out how to calculate each one of these derivatives. So if we try to find easy, do we see that we have the ZX which is a function, so as these a function of X and X as a function of you. So what we remember is we have to do the derivative of C with respect to X, multiplied by the derivative of X with respect to you. And then we add this to the other side which is DZ dy time's D why do you something like this? And those are our only options. So for here we just calculate these derivatives as we see here. So dizzy. D x is going to be equal to two times X plus wider. The third multiplied by D X D U. Which is gonna be our function here, which is just gonna be V squared next B squared. Next we have DZ dy so does any Y here is this going to be three X Y squared And we leave X as a constant and we multiply this by d Y D U. But in this case is just one. And so our answer for this derivative is just two X plus Y to the third multiplied by V squared plus three X Y squared. And we want to find this For you equally to vehicle to one W equal to zero. So all we do is we plug in these answers now just thinking about this quickly. We can also see that X and y are part of our values here. So we can find out what X is equal to what why is equal to at these values. So X would be equal to at these values. We would have to For you times one squared Plus 0 to the 3rd. So this means that X is going to be equal to two and then we have that, why is going to be equal to two plus one, E to the zero E 00 to one. And so we have that, this is equal to three. And so what we do now is we plug in these values so We have that this is going to be equal to two times X which is going to be too Plus 3 to the 3rd V here is equal to one plus three times two times three squared. And that's all we get. So here we would get four plus 27 plus three times to six times 9. And so we would get 31 plus 54 Which is equal to 85. Yeah, there we go. So this is kind of our problem that we do this, we do this every single time. We will do this again for D C D. U. And D C D. Um uh V D C D. W. And we would get the same exact kind of answers again, right? We would get dizzy D W. Um And DZ D V are very similar. Right? In this case it's DZ D X multiplied by dx D W plus DZ dy time's D Y D W. Something like that. And similarly down here we get dizzy. The X time's D X D. V. This time and dizzy. Dy multiplied by the X. Sorry, Dy D D D V. That's all we see here. Just as a reminder that these are the same. So you can keep using this exact same thing here. So we know that this will always be two X plus Y. In the third we know that this will always be three X Y squared for each one of these. And similarly here we will then find that dx dw N D E X T V are just the derivatives with respect to X here. So this one would be then two X plus Y to the third as we had last time. Multiply my D X D W. Which is just www squared here. This would be to expose wider than third again for DVD X multiplied by dx tv which is just going to be too U V. And then we do this again for the for the next one which is just uh three X Y squared three X Y squared D. Y. D. W. In this case it's just V. E. To the W. Right? Because each of the W. Is just uh the review of that is the state of the W. And then for V, you would just get this is E to the W. And we plug our numbers and again and get a very similar answer. So here we would plug this in and we get again, this would be 31 W. Though is zero. So this is zero. This part right here as we saw was equal to 54 by playing on numbers in and V. Here, E to the W. Is one E. To the zero. And so this is just equal to 54. Similarly down here for this guy, for D. V. D. V. We again get 31 Two times u times v is two times 2 times one. And then we get plus 54 Times E. To the W. Which is one. And so we would get 31 times four, which is equal to 124 Plus 54 is a 178. And so these are your answers at your exact point? How fast does the change with respect to W. V. And U. Using the chain wolf?


Similar Solved Questions

5 answers
Duaph [ belo shows the level @utves "differentiable funciion I(x (Ihin curves) well @5constrait %x V)V7,} 0 (Ihick circle} Uslng the concepts the orthogonal gradeent thectent ad the method 0l Lagrange mutiplicrs , eslimate Ihe coordinales corresponding the constrained extrema of f(*.y)(13-0.7), (1.3.0Ti115.021 (0.7,1.31.+15,02) (-07,13) (15 42) (07,.131 (15 02) (07,-1 31 (1.5, 01. (0 151. (+15,01.(0, -151 (11.111(1110[1101a11
Duaph [ belo shows the level @utves " differentiable funciion I(x (Ihin curves) well @5 constrait %x V)V7,} 0 (Ihick circle} Uslng the concepts the orthogonal gradeent thectent ad the method 0l Lagrange mutiplicrs , eslimate Ihe coordinales corresponding the constrained extrema of f(*.y) (13-0....
4 answers
CAUt Kaba- Cu ^F(tw)buundedCun "a (sint)5 ,0s0<%G'
CAUt Kaba- Cu ^ F(tw) buunded Cun "a (sint)5 , 0s0<% G'...
5 answers
0/1 pointsPrevious Answers HarMathAp11 6.5,007My NotesAsk Your TeacherA homeowner planning kitchen remodeling can afford $300 monthly payment. How much can the homeowner borrow for years at 6% _ compounded monthly, and still stay within the budget? (Round your answer to the nearest cent:) 4386.30
0/1 points Previous Answers HarMathAp11 6.5,007 My Notes Ask Your Teacher A homeowner planning kitchen remodeling can afford $300 monthly payment. How much can the homeowner borrow for years at 6% _ compounded monthly, and still stay within the budget? (Round your answer to the nearest cent:) 4386.3...
5 answers
CrEa foilbwing balanced rencdon:NHANO,(s) + NHA" (oql NOj (oqhbHansampke 0f 0.1245 mol ot NHANO_s) dlssolved In water t0 mnake 50.0 solution: The temperature of the solution decreases from 40.4 "C to 22.7 *C, What is AHrn (in kJ) for the balanced chemical equation as written? The specific heat capacity (C ) of the solution is 4.18 Mle "Cl"370UJ"3.700+3699 kJ0 +29J10 -297rJ
crEa foilbwing balanced rencdon: NHANO,(s) + NHA" (oql NOj (oqh bHan sampke 0f 0.1245 mol ot NHANO_s) dlssolved In water t0 mnake 50.0 solution: The temperature of the solution decreases from 40.4 "C to 22.7 *C, What is AHrn (in kJ) for the balanced chemical equation as written? The speci...
5 answers
(1 point) Evaluate the definite integral.0 Jr 12 dc
(1 point) Evaluate the definite integral. 0 Jr 12 dc...
5 answers
The number line has several points labeled. Find the distance between each pair of points. See Example $5 .$$A$ and $C$
The number line has several points labeled. Find the distance between each pair of points. See Example $5 .$ $A$ and $C$...
1 answers
Suppose $|\langle u, v\rangle|=\|u\|\|v\|$. (That is, the Cauchy-Schwarz inequality reduces to an equality.) Show that $u$ and $v$ are linearly dependent.
Suppose $|\langle u, v\rangle|=\|u\|\|v\|$. (That is, the Cauchy-Schwarz inequality reduces to an equality.) Show that $u$ and $v$ are linearly dependent....
5 answers
B: Determine the Formula of an Unknown Hydrate Unkrown sample number: Jhk "Iastx Molecular formula of the unknown: Ho Mass of crucible + unknown hydrate 3586 8 _ 2. Mass of crucible 3869x4n4 3. Mass of unknown hydrate( 1 2) 4. Mass of crucible + anhydrous salt after first heating 35.5Los b. after second heating 3545 /xaL5_ Mass of unknown anhydrous salt (4b 2)6. Mass of water lost (14b)
B: Determine the Formula of an Unknown Hydrate Unkrown sample number: Jhk "Iastx Molecular formula of the unknown: Ho Mass of crucible + unknown hydrate 3586 8 _ 2. Mass of crucible 3869x4n4 3. Mass of unknown hydrate( 1 2) 4. Mass of crucible + anhydrous salt after first heating 35.5Los b. aft...
5 answers
Solve/Compute Surface Vector Integral where S is a triangle with vertex at (1,0,0) (0,1,0) and(0,0,1).With an upward normal orientation.
Solve/Compute Surface Vector Integral where S is a triangle with vertex at (1,0,0) (0,1,0) and (0,0,1). With an upward normal orientation....
5 answers
A stereo speaker is placed between two observers who are 35 mapart, along the line connecting them. If one observer records anintensity level of 61 dB, and the other records an intensity levelof 76 dB, how far is the speaker from each observer?Closer observer ____________m farther observer ____________m
A stereo speaker is placed between two observers who are 35 m apart, along the line connecting them. If one observer records an intensity level of 61 dB, and the other records an intensity level of 76 dB, how far is the speaker from each observer? Closer observer ____________m farther observer ____...
5 answers
A key concept in microscopy is resolution. The resolutionindicates how detailed an image will be. Low resolution gives aclear image, high resolution gives a detailed image. The degree ofresolution you get in an image depends on the microscopic techniqueyou use.a) Give an example of a question where you need a very highresolution in the image to be able to answer the question.b) What type of microscope should be used for this?
A key concept in microscopy is resolution. The resolution indicates how detailed an image will be. Low resolution gives a clear image, high resolution gives a detailed image. The degree of resolution you get in an image depends on the microscopic technique you use. a) Give an example of a question w...
5 answers
Let X be the number of cars sold by a dealership in one day. Thepmf p(x) of X is given in the table. X| 0 1 2 3 4 5 6p(x)| 0.10 0.10 0.20 0.10 0.15 0.15 0.20Find the probability that(a) At least 4 cars are sold.(b) An odd number of cars is sold.
Let X be the number of cars sold by a dealership in one day. The pmf p(x) of X is given in the table. X| 0 1 2 3 4 5 6 p(x)| 0.10 0.10 0.20 0.10 0.15 0.15 0.20 Find the probability that (a) At least 4 cars are sold. (b) An odd numbe...
5 answers
Find the Laplace Transform of the periodic function f(t) using Step functions sin t, 0 <t < T f(t) = f(t) f(t + 2t) (5 marks) 2, T <t < 2t(b) Use Laplace Transform to solve the initial-value problem y" _y' = f(t 5) ,y(0) = 1,y'(0) = l~~ ~n (5 marks)
Find the Laplace Transform of the periodic function f(t) using Step functions sin t, 0 <t < T f(t) = f(t) f(t + 2t) (5 marks) 2, T <t < 2t (b) Use Laplace Transform to solve the initial-value problem y" _y' = f(t 5) ,y(0) = 1,y'(0) = l~~ ~n (5 marks)...
5 answers
PeriodQuarter Year GDP (in billions) 1 1 2015 116.76 2 2 2015 118.26 3 3 2015 119.56 4 2015 119.53 5 1 2016 118.64 6 2 2016 120.09 7 3 2016 119.86 8 4 2016 121.77 9 1 2017 122.5 10 2 2017 123.6 11 3 2017 123.64 12 4 2017 125.11 13 1 2018 126.13 14 2 2018 127.9 15 3 2018 128.48 16 4 2018 129.81 17 1 2019 18 2 2019
Period Quarter Year GDP (in billions) 1 1 2015 116.76 2 2 2015 118.26 3 3 2015 119.56 4 2015 119.53 5 1 2016 118.64 6 2 2016 120.09 7 3 2016 119.86 8 4 2016 121.77 9 1 2017 122.5 10 2 2017 123.6 11 3 2017 123.64 12 4 2017 125.11 13 1 2018 126.13 14 2 2018 127.9 15 3 2018 128.48 16 4 2018 129.81 17 1...
5 answers
[Compute L |e-t f& e% cost cos3t )dt/1
[Compute L |e-t f& e% cost cos3t )dt/1...

-- 0.021390--